Suppose you have the unit interval $[0,1]$. For the first iteration you remove the segment $(1/5,3/5)$. So you are left with two intervals of lengths $1/5$ and $2/5$. You now repeat the process on the remaining intervals to get a set $F$. What is the Hausdorff dimension of this set and can you prove this rigorously.
I think you should note that in level $k$ (starting counting at with $[0,1]$) one has 2^k intervals. The minimum length of the interval is $(\frac{1}{5})^k$ and the maximum length fo an interval is $(\frac{2}{5})^k$
I am having problem with this question as the usual method of solving $\sum c_i^s=1$ doesnt work.
This links to my previous question which is on the same topic.
You have a set that consists of two copies of itself - one scaled by the factor $1/5$ and one scaled by the factor $2/5$. The similarity dimension is the unique number $s$ satisfying $$(1/5)^s + (2/5)^s = 1.$$ Again, this characterizes the dimension uniquely, though doesn't express it in closed form. In general, an equation of this type can be solved in closed form only when the scaling factors are exponentially commensurable. I provide some examples in my answer to this question.
Of course, we can compute your dimension as closely as we want - it is approximately $0.56389552425993647949$.